3.49.72 \(\int \frac {e^x (-2-2 x+2 x^2+2 x^3)+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} (e^x (2+8 x+10 x^2+4 x^3)+e^{e^x} (-5 e^x+e^{2 x} (-5 x-5 x^2)))}{8 x^2+16 x^3+8 x^4+e^{\frac {2 (5 e^{e^x} x-2 x^2-2 x^3)}{2+2 x}} (8+16 x+8 x^2)+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} (16 x+32 x^2+16 x^3)} \, dx\)

Optimal. Leaf size=33 \[ \frac {e^x}{4 \left (e^{x \left (-x+\frac {5 e^{e^x}}{2 (1+x)}\right )}+x\right )} \]

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Rubi [F]  time = 145.38, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^x \left (-2-2 x+2 x^2+2 x^3\right )+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} \left (e^x \left (2+8 x+10 x^2+4 x^3\right )+e^{e^x} \left (-5 e^x+e^{2 x} \left (-5 x-5 x^2\right )\right )\right )}{8 x^2+16 x^3+8 x^4+e^{\frac {2 \left (5 e^{e^x} x-2 x^2-2 x^3\right )}{2+2 x}} \left (8+16 x+8 x^2\right )+e^{\frac {5 e^{e^x} x-2 x^2-2 x^3}{2+2 x}} \left (16 x+32 x^2+16 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(-2 - 2*x + 2*x^2 + 2*x^3) + E^((5*E^E^x*x - 2*x^2 - 2*x^3)/(2 + 2*x))*(E^x*(2 + 8*x + 10*x^2 + 4*x^3
) + E^E^x*(-5*E^x + E^(2*x)*(-5*x - 5*x^2))))/(8*x^2 + 16*x^3 + 8*x^4 + E^((2*(5*E^E^x*x - 2*x^2 - 2*x^3))/(2
+ 2*x))*(8 + 16*x + 8*x^2) + E^((5*E^E^x*x - 2*x^2 - 2*x^3)/(2 + 2*x))*(16*x + 32*x^2 + 16*x^3)),x]

[Out]

-1/4*Defer[Int][E^(x + 2*x^2)/(E^((5*E^E^x*x)/(2 + 2*x)) + E^x^2*x)^2, x] - (5*Defer[Int][E^(E^x + 2*x + 2*x^2
)/(E^((5*E^E^x*x)/(2 + 2*x)) + E^x^2*x)^2, x])/8 + (5*Defer[Int][(E^(E^x + 2*x + 2*x^2)*x)/(E^((5*E^E^x*x)/(2
+ 2*x)) + E^x^2*x)^2, x])/8 - Defer[Int][(E^(x + 2*x^2)*x^2)/(E^((5*E^E^x*x)/(2 + 2*x)) + E^x^2*x)^2, x]/2 - (
5*Defer[Int][E^(E^x + x + 2*x^2)/((1 + x)^2*(E^((5*E^E^x*x)/(2 + 2*x)) + E^x^2*x)^2), x])/8 + (5*Defer[Int][E^
(E^x + x + 2*x^2)/((1 + x)*(E^((5*E^E^x*x)/(2 + 2*x)) + E^x^2*x)^2), x])/8 + (5*Defer[Int][E^(E^x + 2*x + 2*x^
2)/((1 + x)*(E^((5*E^E^x*x)/(2 + 2*x)) + E^x^2*x)^2), x])/8 + Defer[Int][E^(x + x^2)/(E^((5*E^E^x*x)/(2 + 2*x)
) + E^x^2*x), x]/4 - (5*Defer[Int][E^(E^x + 2*x + x^2)/(E^((5*E^E^x*x)/(2 + 2*x)) + E^x^2*x), x])/8 + Defer[In
t][(E^(x + x^2)*x)/(E^((5*E^E^x*x)/(2 + 2*x)) + E^x^2*x), x]/2 - (5*Defer[Int][E^(E^x + x + x^2)/((1 + x)^2*(E
^((5*E^E^x*x)/(2 + 2*x)) + E^x^2*x)), x])/8 + (5*Defer[Int][E^(E^x + 2*x + x^2)/((1 + x)*(E^((5*E^E^x*x)/(2 +
2*x)) + E^x^2*x)), x])/8

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{x+x^2} \left (-5 e^{e^x+\frac {5 e^{e^x} x}{2+2 x}}-5 e^{e^x+x+\frac {5 e^{e^x} x}{2+2 x}} x (1+x)+2 e^{x^2} (-1+x) (1+x)^2+2 e^{\frac {5 e^{e^x} x}{2+2 x}} (1+x)^2 (1+2 x)\right )}{8 (1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx\\ &=\frac {1}{8} \int \frac {e^{x+x^2} \left (-5 e^{e^x+\frac {5 e^{e^x} x}{2+2 x}}-5 e^{e^x+x+\frac {5 e^{e^x} x}{2+2 x}} x (1+x)+2 e^{x^2} (-1+x) (1+x)^2+2 e^{\frac {5 e^{e^x} x}{2+2 x}} (1+x)^2 (1+2 x)\right )}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx\\ &=\frac {1}{8} \int \left (\frac {e^{x+x^2} \left (2-5 e^{e^x}+8 x-5 e^{e^x+x} x+10 x^2-5 e^{e^x+x} x^2+4 x^3\right )}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )}-\frac {e^{x+2 x^2} \left (2+4 x-5 e^{e^x} x+6 x^2-5 e^{e^x+x} x^2+8 x^3-5 e^{e^x+x} x^3+4 x^4\right )}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2}\right ) \, dx\\ &=\frac {1}{8} \int \frac {e^{x+x^2} \left (2-5 e^{e^x}+8 x-5 e^{e^x+x} x+10 x^2-5 e^{e^x+x} x^2+4 x^3\right )}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )} \, dx-\frac {1}{8} \int \frac {e^{x+2 x^2} \left (2+4 x-5 e^{e^x} x+6 x^2-5 e^{e^x+x} x^2+8 x^3-5 e^{e^x+x} x^3+4 x^4\right )}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx\\ &=-\left (\frac {1}{8} \int \left (\frac {2 e^{x+2 x^2}}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2}+\frac {4 e^{x+2 x^2} x}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2}-\frac {5 e^{e^x+x+2 x^2} x}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2}+\frac {6 e^{x+2 x^2} x^2}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2}-\frac {5 e^{e^x+2 x+2 x^2} x^2}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2}+\frac {8 e^{x+2 x^2} x^3}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2}-\frac {5 e^{e^x+2 x+2 x^2} x^3}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2}+\frac {4 e^{x+2 x^2} x^4}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2}\right ) \, dx\right )+\frac {1}{8} \int \left (\frac {2 e^{x+x^2}}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )}-\frac {5 e^{e^x+x+x^2}}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )}+\frac {8 e^{x+x^2} x}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )}-\frac {5 e^{e^x+2 x+x^2} x}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )}+\frac {10 e^{x+x^2} x^2}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )}-\frac {5 e^{e^x+2 x+x^2} x^2}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )}+\frac {4 e^{x+x^2} x^3}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )}\right ) \, dx\\ &=-\left (\frac {1}{4} \int \frac {e^{x+2 x^2}}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx\right )+\frac {1}{4} \int \frac {e^{x+x^2}}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )} \, dx-\frac {1}{2} \int \frac {e^{x+2 x^2} x}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx-\frac {1}{2} \int \frac {e^{x+2 x^2} x^4}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx+\frac {1}{2} \int \frac {e^{x+x^2} x^3}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )} \, dx+\frac {5}{8} \int \frac {e^{e^x+x+2 x^2} x}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx+\frac {5}{8} \int \frac {e^{e^x+2 x+2 x^2} x^2}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx+\frac {5}{8} \int \frac {e^{e^x+2 x+2 x^2} x^3}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx-\frac {5}{8} \int \frac {e^{e^x+x+x^2}}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )} \, dx-\frac {5}{8} \int \frac {e^{e^x+2 x+x^2} x}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )} \, dx-\frac {5}{8} \int \frac {e^{e^x+2 x+x^2} x^2}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )} \, dx-\frac {3}{4} \int \frac {e^{x+2 x^2} x^2}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx+\frac {5}{4} \int \frac {e^{x+x^2} x^2}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )} \, dx-\int \frac {e^{x+2 x^2} x^3}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )^2} \, dx+\int \frac {e^{x+x^2} x}{(1+x)^2 \left (e^{\frac {5 e^{e^x} x}{2+2 x}}+e^{x^2} x\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.63, size = 38, normalized size = 1.15 \begin {gather*} \frac {e^{x+x^2}}{4 \left (e^{\frac {5 e^{e^x} x}{2 (1+x)}}+e^{x^2} x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-2 - 2*x + 2*x^2 + 2*x^3) + E^((5*E^E^x*x - 2*x^2 - 2*x^3)/(2 + 2*x))*(E^x*(2 + 8*x + 10*x^2 +
4*x^3) + E^E^x*(-5*E^x + E^(2*x)*(-5*x - 5*x^2))))/(8*x^2 + 16*x^3 + 8*x^4 + E^((2*(5*E^E^x*x - 2*x^2 - 2*x^3
))/(2 + 2*x))*(8 + 16*x + 8*x^2) + E^((5*E^E^x*x - 2*x^2 - 2*x^3)/(2 + 2*x))*(16*x + 32*x^2 + 16*x^3)),x]

[Out]

E^(x + x^2)/(4*(E^((5*E^E^x*x)/(2*(1 + x))) + E^x^2*x))

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fricas [A]  time = 0.57, size = 33, normalized size = 1.00 \begin {gather*} \frac {e^{x}}{4 \, {\left (x + e^{\left (-\frac {2 \, x^{3} + 2 \, x^{2} - 5 \, x e^{\left (e^{x}\right )}}{2 \, {\left (x + 1\right )}}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-5*x^2-5*x)*exp(x)^2-5*exp(x))*exp(exp(x))+(4*x^3+10*x^2+8*x+2)*exp(x))*exp((5*x*exp(exp(x))-2*x
^3-2*x^2)/(2*x+2))+(2*x^3+2*x^2-2*x-2)*exp(x))/((8*x^2+16*x+8)*exp((5*x*exp(exp(x))-2*x^3-2*x^2)/(2*x+2))^2+(1
6*x^3+32*x^2+16*x)*exp((5*x*exp(exp(x))-2*x^3-2*x^2)/(2*x+2))+8*x^4+16*x^3+8*x^2),x, algorithm="fricas")

[Out]

1/4*e^x/(x + e^(-1/2*(2*x^3 + 2*x^2 - 5*x*e^(e^x))/(x + 1)))

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-5*x^2-5*x)*exp(x)^2-5*exp(x))*exp(exp(x))+(4*x^3+10*x^2+8*x+2)*exp(x))*exp((5*x*exp(exp(x))-2*x
^3-2*x^2)/(2*x+2))+(2*x^3+2*x^2-2*x-2)*exp(x))/((8*x^2+16*x+8)*exp((5*x*exp(exp(x))-2*x^3-2*x^2)/(2*x+2))^2+(1
6*x^3+32*x^2+16*x)*exp((5*x*exp(exp(x))-2*x^3-2*x^2)/(2*x+2))+8*x^4+16*x^3+8*x^2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.14, size = 32, normalized size = 0.97

method	result	size
risch	\(\frac {{\mathrm e}^{x}}{4 x +4 \,{\mathrm e}^{-\frac {x \left (2 x^{2}-5 \,{\mathrm e}^{{\mathrm e}^{x}}+2 x \right )}{2 \left (x +1\right )}}}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((-5*x^2-5*x)*exp(x)^2-5*exp(x))*exp(exp(x))+(4*x^3+10*x^2+8*x+2)*exp(x))*exp((5*x*exp(exp(x))-2*x^3-2*x
^2)/(2*x+2))+(2*x^3+2*x^2-2*x-2)*exp(x))/((8*x^2+16*x+8)*exp((5*x*exp(exp(x))-2*x^3-2*x^2)/(2*x+2))^2+(16*x^3+
32*x^2+16*x)*exp((5*x*exp(exp(x))-2*x^3-2*x^2)/(2*x+2))+8*x^4+16*x^3+8*x^2),x,method=_RETURNVERBOSE)

[Out]

1/4*exp(x)/(x+exp(-1/2*x*(2*x^2-5*exp(exp(x))+2*x)/(x+1)))

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maxima [B]  time = 0.56, size = 161, normalized size = 4.88 \begin {gather*} \frac {{\left (4 \, x^{4} + 8 \, x^{3} + 6 \, x^{2} - 5 \, {\left ({\left (x^{3} + x^{2}\right )} e^{x} + x\right )} e^{\left (e^{x}\right )} + 4 \, x + 2\right )} e^{\left (x^{2} + x + \frac {5 \, e^{\left (e^{x}\right )}}{2 \, {\left (x + 1\right )}}\right )}}{4 \, {\left ({\left (4 \, x^{5} + 8 \, x^{4} + 6 \, x^{3} + 4 \, x^{2} - 5 \, {\left (x^{2} + {\left (x^{4} + x^{3}\right )} e^{x}\right )} e^{\left (e^{x}\right )} + 2 \, x\right )} e^{\left (x^{2} + \frac {5 \, e^{\left (e^{x}\right )}}{2 \, {\left (x + 1\right )}}\right )} + {\left (4 \, x^{4} + 8 \, x^{3} + 6 \, x^{2} - 5 \, {\left ({\left (x^{3} + x^{2}\right )} e^{x} + x\right )} e^{\left (e^{x}\right )} + 4 \, x + 2\right )} e^{\left (\frac {5}{2} \, e^{\left (e^{x}\right )}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-5*x^2-5*x)*exp(x)^2-5*exp(x))*exp(exp(x))+(4*x^3+10*x^2+8*x+2)*exp(x))*exp((5*x*exp(exp(x))-2*x
^3-2*x^2)/(2*x+2))+(2*x^3+2*x^2-2*x-2)*exp(x))/((8*x^2+16*x+8)*exp((5*x*exp(exp(x))-2*x^3-2*x^2)/(2*x+2))^2+(1
6*x^3+32*x^2+16*x)*exp((5*x*exp(exp(x))-2*x^3-2*x^2)/(2*x+2))+8*x^4+16*x^3+8*x^2),x, algorithm="maxima")

[Out]

1/4*(4*x^4 + 8*x^3 + 6*x^2 - 5*((x^3 + x^2)*e^x + x)*e^(e^x) + 4*x + 2)*e^(x^2 + x + 5/2*e^(e^x)/(x + 1))/((4*
x^5 + 8*x^4 + 6*x^3 + 4*x^2 - 5*(x^2 + (x^4 + x^3)*e^x)*e^(e^x) + 2*x)*e^(x^2 + 5/2*e^(e^x)/(x + 1)) + (4*x^4
+ 8*x^3 + 6*x^2 - 5*((x^3 + x^2)*e^x + x)*e^(e^x) + 4*x + 2)*e^(5/2*e^(e^x)))

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mupad [B]  time = 3.75, size = 48, normalized size = 1.45 \begin {gather*} \frac {{\mathrm {e}}^x}{4\,x}-\frac {{\mathrm {e}}^x}{x\,\left (4\,x\,{\mathrm {e}}^{\frac {2\,x^2-5\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}+2\,x^3}{2\,x+2}}+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(2*x^2 - 5*x*exp(exp(x)) + 2*x^3)/(2*x + 2))*(exp(exp(x))*(5*exp(x) + exp(2*x)*(5*x + 5*x^2)) - exp
(x)*(8*x + 10*x^2 + 4*x^3 + 2)) + exp(x)*(2*x - 2*x^2 - 2*x^3 + 2))/(exp(-(2*(2*x^2 - 5*x*exp(exp(x)) + 2*x^3)
)/(2*x + 2))*(16*x + 8*x^2 + 8) + exp(-(2*x^2 - 5*x*exp(exp(x)) + 2*x^3)/(2*x + 2))*(16*x + 32*x^2 + 16*x^3) +
8*x^2 + 16*x^3 + 8*x^4),x)

[Out]

exp(x)/(4*x) - exp(x)/(x*(4*x*exp((2*x^2 - 5*x*exp(exp(x)) + 2*x^3)/(2*x + 2)) + 4))

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sympy [A]  time = 1.13, size = 32, normalized size = 0.97 \begin {gather*} \frac {e^{x}}{4 x + 4 e^{\frac {- 2 x^{3} - 2 x^{2} + 5 x e^{e^{x}}}{2 x + 2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-5*x**2-5*x)*exp(x)**2-5*exp(x))*exp(exp(x))+(4*x**3+10*x**2+8*x+2)*exp(x))*exp((5*x*exp(exp(x))
-2*x**3-2*x**2)/(2*x+2))+(2*x**3+2*x**2-2*x-2)*exp(x))/((8*x**2+16*x+8)*exp((5*x*exp(exp(x))-2*x**3-2*x**2)/(2
*x+2))**2+(16*x**3+32*x**2+16*x)*exp((5*x*exp(exp(x))-2*x**3-2*x**2)/(2*x+2))+8*x**4+16*x**3+8*x**2),x)

[Out]

exp(x)/(4*x + 4*exp((-2*x**3 - 2*x**2 + 5*x*exp(exp(x)))/(2*x + 2)))

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